Could mass just be light moving in another dimension?

Question

Could mass just be perceived as light moving along a geodesic through an additional spatial dimension (either invisible or somehow curled up into itself)? Since the light would be moving in another dimension, to us its velocity will be less than the speed of light, giving the illusion of mass.

Answer 1

In order to illustrate the difficulties associated with such an approach, I will mention an example. One way to obtain a toy model according to your requirement is Kaluza-Klein theory, which assumes space-time is 5-dimensional, and which kind-of obtains Maxwell's equations (classical electrodynamics) from Einstein's field equations in 5D.

The reason I say "kind of" is that it depends on what letters of the alphabet you give the variables in the obtained equations. The mere fact that some equations of a synthetic theory formally correspond to equations that are already known to satisfy experiments, does not automatically imply that this theory is useful. Only if the theory does not "predict" a lot of quite arbitrary things that have never been observed, AND it predicts the things that are already known, it can be considered useful.

To give you a quick access to what Kaluza-Klein-theory predicts here, let's consider the special case where the 5D spacetime is (approximately) completely flat (i.e. there is no curvature/gravity) and one of its dimensions is "rolled up" on a small scale. You can imagine the 5th dimension together with some other spatial dimension (e.g. "x") as a piece of paper, which you literally roll up to a tube with your hands. Since this nowhere requires the paper to stretch or compress, the tube remains geometrically flat.

Then, rolling up the 5th dimension is equivalent to having a dimension that has finite extent and periodic boundary conditions (due to the tube-like topology). If you want to explore what happens to a (massless) wave that propagates in this 5D spacetime, you could for example write down the d'Alembert equation for a wave function $\psi$, which can be considered representing one component of electromagnetic waves (e.g. the electric field strength in x-direction): $$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}-\frac{\partial^2 \psi}{\partial x^2}=\frac{\partial^2 \psi}{\partial w^2}$$ where $w$ shall denote the coordinate in the 5th dimension, and $x$ may represent one of the usual macroscopic spatial dimensions. The fact that the $w$-term on the right-hand side has positive sign (it would have obtained a negative sign as well if we had written it on the left-hand side) indicates that this is an additional spatial dimension (as opposed to an additional temporal dimension).

Now, due to the periodic boundary conditions, the coordinate value $w+L$ is equivalent to $w$ where $L$ is the circumference of the rolled-up 5th dimension. For the wave function $\psi$ this means $$\psi(w+L)=\psi(w)$$ A function with this periodic property can generally be written as a Fourier series: $$\psi(w)=\sum_{k=-\infty}^\infty \psi_k \cdot \exp(i 2\pi k w/L)$$ Consider only one component of this Fourier series, for example $$\psi(w)=\psi_{k_0} \cdot \exp(i 2\pi k_0 w/L)$$ Then the second order derivative on the right-hand side of the wave equation becomes $$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}-\frac{\partial^2 \psi}{\partial x^2}=\frac{\partial^2 \psi}{\partial w^2}=-\frac{4\pi^2 k_0^2}{L^2}\psi$$ The factor on the right-hand side before $\psi$ is what is called an eigenvalue (in this case of the second order derivative under the assumed periodic boundary conditions). Sounds good, eigenvalues... quantum mechanics... seems we are on the right track. So, let's see where this approach can carry us with respect to mass.

Up to now, we have only considered massless particles ("light"). One of the quantum-mechanical equations that might describe massive particles relativistically is Klein-Gordon equation, which is $$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2}-\frac{\partial^2 \psi}{\partial x^2}=-\frac{m^2c^2}{\hbar^2}\psi$$ If we compare our massless wave equation in 5D with the Klein-Gordon equation, it seems pretty obvious that we "should" identify the mass term according to $$\frac{m^2c^2}{\hbar^2}=\frac{4\pi^2 k_0^2}{L^2}$$ or finally $$m=k_0\frac{h}{cL}$$ We have identified mass as resulting from the finite circumference of a rolled up dimension! And moreover, it does what you think it should do: if you carefully consider what the above Fourier component part of the wave function does, you will see that it rotates in the fifth dimension, and if there is also motion in the other spatial dimensions, this circular motion will become a spiralling motion at the speed of light in 5D (since we have started with a wave equation, there is no doubt about this). If we look at this spiralling motion macroscopically, we will be blind for the tiny extra dimension and see only the averaged motion in the ordinary dimensions, which will have a speed $<c$.

Another interesting point about this is that $L$ can be considered the Compton wavelength of a particle with mass $m$, at least for $k_0=1$. If that doesn't sound interesting, then what does? But wait, $k_0=1$? And what about $k_0=2,3,\dots$? If $k_0=1$ shall represent the electron (and probably $k_0=-1$ the positron, yeah!), then what do the higher integer numbers represent? Duh! There is simply no particle with twice or three times the mass of the electron!

This integer spectrum of an infinite number of masses/particles is called the Kaluza-Klein tower. It illustrates that a naive interpretation of seemingly corresponding equations in toy models can be problematic. As a theoretical physicist, you have the responsibility to assign a meaning to every quantity in a new model, that allows it to be tested against experiments.

And the problems don't even stop at the Kaluza-Klein tower. Next question is, why there exists a rolled-up dimension of length $L$ in the first place? Well that may probably be explained by boundary conditions of the universe around us ("tube in, tube out"). But electrons are Fermions, and don't even satisfy Klein-Gordon equation (like assumed above), but rather Dirac's equation. And moreover, we have supposed that the wave, that propagates into the 5th dimension and thereby gets mass, is light, but light (photons) is actually massless in reality. At least there, we might try to weasel out by saying that the photon is represented by the component $k_0=0$ which is massless and hence, always travels at the speed of light. Then, finally, we have more than one known elementary particle besides the electron, so this is probably not making it easier to model all of them by extra dimensions and simultaneously answer all of the above questions.

And so on and so forth. That is why, as of today, nobody can claim to have found something better than the standard model, where masses (or rather coupling constants) are more or less designed-in from the start.

Answer 2

Let $P^M= (p^\mu, m)$ where $\mu$ is a spacetime index in $d$ dimensions and $M$ is in $d+1$ dimensions. If $P$ is a lightlike momentum vector then $P^2 = p^2 - m^2 = 0$, and we see that $p^2=m^2$ is an on-shell massive momentum in $d$ dimensions. (Maybe your question came from this observation?)

Using a higher dimensional lightlike momentum to produce a massive momentum in this way is a standard technique in mathematical physics, and is referred to as dimensional reduction. For example, it can make calculations more simple to phrase massive $d$-dimensional problems as massless higher dimensional ones. My experience is in the study of scattering amplitudes in QFT, where amplitudes for massless particles are easier to work with.

The other answers addressed whether this is a good model for nature, so I won't comment on this. I thought it would be relevant to note that the technique is still used regardless of physical interpretation.

Edit; To illustrate how my first paragraph relates to the question more clearly, let $v^\mu = (v^0, v^i)$ be a timelike 4-vector, in metric signature $(+---)$. Then we see that $$v^2 = (v^0)^2 - \sum_{i=1}^3(v^i)^2 = c^2.$$ If we define $v^4 = c$ then we can move the $c^2$ to the other side and write that $$(v^0)^2 - \sum_{i=1}^4(v^i)^2 = 0,$$ which looks like the on-shell relation $V^2=0$ for a massless 5-vector with components $V^M = (v^0, v^i,c)$ in metric signature $(+----)$. (Multiply by $m$ and go to natural units with $c=1$ to recover my statement in terms of momenta)

Answer 3

These dimensions cant be big, as we would notice the effect on planet orbits for example. Force decreases with distance as a power that is one less than an amount of dimensions.

https://en.m.wikipedia.org/wiki/Inverse-square_law

These dimensions cant be on the scale of existing particles, as we would notice missing energy in our collider experiments, if particles would be pushed off in these dimensions while they pass through the detector.

These dimensions can be significantly smaller, we would have no way of detecting the difference.

https://en.m.wikipedia.org/wiki/String_theory

But then if we cant reliably say anything about is, is it really a good idea to start the physics interest with such a topic?

https://www.scientificamerican.com/article/is-string-theory-science/